(a) Field of the invention
The present invention relates to a method for detecting a line-to-ground fault location in power network, and more particularly, detecting the line-to-ground fault location with direct 3-phase circuit analysis without using a symmetrical component transformation, whereby even in an unbalanced 3-phase circuit the line-to-ground fault location can be accurately detected.
b) Description of the Related Art
Rapid growth of economy has resulted in large scale of power systems, and an excessive increase in transmission and distribution networks of electric power systems causes many kinds of faults by various causes. Transmission and distribution networks of electric power systems are playing very important roles as the links between the power suppliers and the consumers. However, because most of lines are exposed to air, lightning, contact of animals or mal-functioning of protection devices causes many kinds of faults. When a line-to-ground fault occurs, detecting a fault location rapidly and precisely separating the part of the network including the fault location from the rest part of the network until repairing the fault being finished is very important to minimize power interruption rate and to provide highly reliable power supplying service and electric power of high quality.
Transformation of 3-phase networks to symmetrical component systems (symmetrical component transformation) is generally used in conventional methods for detecting the line-to-ground fault locations. A 3-phase balanced network can be transformed to a symmetrical component system, which has no coupling between sequences so that the systems of equation may be solved easily. In other words, diagonal sequence impedance matrices are obtained in case of the balanced networks, thus sequence voltage and current can be expressed without any coupling among the sequences ([1] G. B. Ancell, N. C. Pahalawaththa, xe2x80x9cMaximum Likelyhood Estimation of Fault Location on Transmission Lines Using Travelling Wavesxe2x80x9d, IEEE Transactions on PWRD, Vol. 9, No. 9, 1994, pp. 680-689; [2] T. Takagi, Y. Yamakoshi, J. Baba, K. Uemura, T. Sakaguchi, xe2x80x9cA New Algorithm of an Accurate Fault Location for EHV/UHV Transmission Lines: Partxe2x80x94Fourier Transform Methodxe2x80x9d, IEEE Trans. on PAS, Vol. PAS-100, No. 3, 1988, pp. 1316-1323; and [3] M. S. Sachdev, R. Agarwal, xe2x80x9cA Technique For Estimating Transmission Line Fault Locations form Digital Impedance Relay Measurementsxe2x80x9d, IEEE Trans. on PWRD, Vol. 3, No.1, 1988, pp 121-129).
Following is a detailed description of an example of the conventional method for detecting the line to ground fault location in a power system using the symmetrical component transformation. FIG. 1 shows a simplified diagram of a balanced 3-phase power network.
Assuming the capacitance in this network is so small that it is negligible, A-phase voltage at the measuring point (relay) is given as:
VSa=(1xe2x88x92d)Zl1xc3x97(ISa+kIS0)+IfRf xe2x80x83xe2x80x83(1) 
where, VSa: a-phase voltage; ISa: a-phase current; IS0: zero sequence current; Zl0,Zl: zero and positive sequence impedance of line; kxe2x89xa1(Zl0xe2x88x92Zl1)/Z1; If: fault current; Rf: fault resistance; and d: fault distance from the other end.
In equation (1), all the impedances except fault resistance are known values. In addition, the phase voltage and current are also available from measurement. One of two unknown variables, a fault current can be obtained using the current distribution factor and another one, a fault resistance can be eliminated during derivation of fault location equation, whose details are described in the following.
For a single line-to-ground fault, the following relation holds: If=3If2, and the negative sequence component of the fault current If2 can be obtained from the distribution factor, Df=If2/IS2 and the negative sequence current (IS2) at the measuring point.
FIG. 2 is a negative sequence network obtained by symmetrical component transformation of a faulted network given in FIG. 1.
The negative sequence circuit for the balanced system in FIG. 2 is shown in FIG. 3. Applying KVL to the route A, the following voltage equation is obtained:
ZS2IS2+(1xe2x88x92d)Zl2IS2xe2x88x92dZl2Ir2xe2x88x92Zr2Ir2=0 xe2x80x83xe2x80x83(2) 
where, ZS2,Zr2: equivalent negative sequence impedances for the source and load; IS2, Ir2: negative sequence current flowing into the fault from the source and load side; and Zl2: negative sequence line impedance.
From Equation (2), distribution factor Db can be obtained.                               D          b                =                                            I              r2                                      I              S2                                =                                                                                          (                                          1                      -                      d                                        )                                    ⁢                                      Zl                    2                                                  +                                  Z                  S2                                                                              dZl                  2                                +                                  Z                  r2                                                      =                                                            -                                      dA                    1                                                  +                                  B                  1                                                                              dA                  1                                +                                  D                  1                                                                                        (        3        )            
where, A1=Zl2, B1=ZS2+Zl2 and D1=Zr2.
Then a negative sequence current distribution factor Df is obtained using Eq. (3):                               D          f                =                                            I              f2                                      I              S2                                =                                                                      I                  S2                                +                                  I                  r2                                                            I                S2                                      =                                                            B                  1                                +                                  D                  1                                                                              dA                  1                                +                                  D                  1                                                                                        (        4        )            
Equation (4) provides a way to obtain negative sequence component of the fault current from the source current.
Substituting If in Equation (1) by If=3If2 and Equation (4), we obtain:                                                                         V                Sa                            =                              xe2x80x83                            ⁢                                                                    (                                          1                      -                      d                                        )                                    ⁢                                      Zl                    1                                    xc3x97                                      (                                                                  I                        Sa                                            +                                              kI                        S0                                                              )                                                  +                                  3                  ⁢                                      I                    f2                                    ⁢                                      R                    f                                                                                                                          =                              xe2x80x83                            ⁢                                                                    (                                          1                      -                      d                                        )                                    ⁢                                      Zl                    1                                    xc3x97                                      (                                                                  I                        Sa                                            +                                              kI                        S0                                                              )                                                  +                                                                            3                      ⁢                                              (                                                                              B                            1                                                    +                                                      D                            1                                                                          )                                            ⁢                                              I                        S2                                                                                                            dA                        1                                            +                                              D                        1                                                                              ⁢                                                            R                      f                                        .                                                                                                          (        5        )            
Rearrangement of Equation (5) yields:
(VSaxe2x88x92(1xe2x88x92d)Zl1(ISa+kIS0))(dA1+D1)+3(B1D1)IS2Rf=0 xe2x80x83xe2x80x83(6) 
Equation (6) is rearranged again to form a second order polynomial with respect to fault distance variable d as follows:
d2(ar+jai)+d(br+jbi)+(crjc1)+Rf(dr+jd1)=0 xe2x80x83xe2x80x83(7) 
where, ar+jai=(ISa+kIS0)Zl1A1,
br+jbi=(ISa+kIS0)Zl1D1+(VSaxe2x88x92((ISa+kIS0)Zl1))A1,
cr+jci=(VSa+kIS0)Zl1))D1,
dr+jdi=3(B1+D1)IS2.
From the imaginary part of Equation (7), the fault resistance is obtained.
xe2x88x92(d2ai+dbici)/di=Rf xe2x80x83xe2x80x83(8) 
Replacing Rf in the real part of Equation (7) by Equation (8),                                                         d              2                        ⁡                          (                                                a                  r                                -                                                                            d                      r                                                              d                      1                                                        ⁢                                      a                    1                                                              )                                +                      d            ⁡                          (                                                b                  r                                -                                                                            d                      r                                                              d                      1                                                        ⁢                                      b                    1                                                              )                                +                      c            r                    -                                                    d                r                                            d                1                                      ⁢                          c              1                                      =        0                            (                  8          ⁢                      -1)                              
is obtained. Then the fault distance 1xe2x88x92d can be obtained by solving Equation (8-1).
The advantage of the above method is that it can be easily applied to a balanced network. Zero sequence, positive sequence and negative sequence can be easily analyzed because they are not correlated, that is, there is no couplings, or equivalently, mutual impedances among the sequences. In the above example, the fault location can be simply determined by analyzing the negative sequence circuit only.
However, the above conventional method can only be applied to a balanced network, because the simplified equations of zero sequence, positive sequence and negative sequence, which are not correlated, can be obtained by symmetrical component transformation only in the balanced network. Thus, the conventional method is not available for unbalanced systems.
The present invention has been made in an effort to solve the above problems.
It is an object of the present invention to provide a method for detecting a line-to-ground fault location in a power network including a fault resistance, not using the symmetrical component transformation but using direct 3-phase circuit analysis. The line-to-ground fault location detecting method of the present invention requires voltage and current of a fundamental frequency measured at a relay.
In the method using direct 3-phase circuit analysis of the present invention, matrix inverse lemma is applied to simplify matrix inversion calculations, thus the line-to-ground fault location can be easily and accurately determined even in the case of an unbalanced network without using the symmetrical component transformation.
By using the method of the present invention, the line-to-ground fault location can be directly analyzed whether the 3-phase network is balanced or unbalanced.
To achieve the above object, the present invention provides a method for detecting a line-to-ground fault location in a power network comprising the steps of:
determining elements of a line impedance matrix and phase voltage and current at a relay;
determining a line-to-ground fault distance by substituting said elements of line impedance matrix, said phase voltage and current into a fault location equation based on direct circuit analysis;
wherein said fault location equation is derived from a model consisting of said phase voltage and current at the relay, a fault current, a fault resistance and the line-to-ground fault distance;
wherein the model is described by the equation: VSabc=(1xe2x88x92d)xc3x97Zlabcxc3x97ISabc+Vfabc,
where, VSabc=[VsaVSbVSc]xe2x80x2: phase voltage vector, ISabc=[ISaISbISc]xe2x80x2: phase current vector, Vfabc=[VfaVfbVfc]xe2x80x2: phase voltage vector at the fault location,             Zl      abc        =                  [                                                            Zl                aa                                                                    Zl                ab                                                                    Zl                                  a                  ⁢                                      xe2x80x83                                    ⁢                  c                                                                                                        Zl                ba                                                                    Zl                bb                                                                    Zl                bc                                                                                        Zl                ca                                                                    Zl                cb                                                                    Zl                cc                                                    ]            ⁢              :            ⁢              xe2x80x83            ⁢      line      ⁢              xe2x80x83            ⁢      impedance      ⁢              xe2x80x83            ⁢      matrix        ,
xe2x80x83If: fault current, 1xe2x88x92d: fault distance;
wherein said fault location equation is derived by using the matrix inverse lemma: (Axe2x88x921+BCD)xe2x88x921=Axe2x88x92AB(Cxe2x88x921+DAB)xe2x88x921DA, to simplify an inverse matrix calculation; and
wherein the fault location equation is derived by direct circuit analysis without using the conventional symmetrical component transformation method.